A wire having length l carries a current i directed along the x...

The magnetic force acting on a current-carrying wire is given by $\vec{F} = i(\vec{l} \times \vec{B})$.

Since the wire is directed along the $x$-axis, the length vector is $\vec{l} = l \hat{i}$.

The magnetic field is given as $\vec{B} = B_0 (\hat{i} + \hat{j} + \hat{k})$.

Substituting these into the force formula:

$\vec{F} = i (l \hat{i} \times B_0 (\hat{i} + \hat{j} + \hat{k}))$

$\vec{F} = i l B_0 (\hat{i} \times \hat{i} + \hat{i} \times \hat{j} + \hat{i} \times \hat{k})$

Using the properties of cross products, $\hat{i} \times \hat{i} = 0$, $\hat{i} \times \hat{j} = \hat{k}$, and $\hat{i} \times \hat{k} = -\hat{j}$, we get:

$\vec{F} = i l B_0 (\hat{k} - \hat{j})$

The magnitude of the magnetic force is:

$|\vec{F}| = i l B_0 \sqrt{(-1)^2 + (1)^2} = \sqrt{2} B_0 i l$